Chapter 23 Electric Fields CHAPTER OUTLINE 23.3 Coulomb’s Law 23.4 The Electric Field 23.6 Electric Field Lines 23.7 Motion of Charged Particles in a Uniform Electric Field

electric charge has the following important properties: There are two kinds of charges in nature; charges of opposite sign attract one Properties of electric charge another and charges of the same sign repel one another. Total charge in an isolated system is conserved. Charge is quantized Q=nq.

23.3 Coulomb’s Law Charles Coulomb (1736–1806) measured the magnitudes of the electric forces between charged objects using the torsion balance

23.3 Coulomb’s Law From Coulomb’s experiments, we can generalize the following properties of the electric force between two stationary charged particles. The electric force is inversely proportional to the square of the separation r between the particles and directed along the line joining them; is proportional to the product of the charges q1 and q2 on the two particles; is attractive if the charges are of opposite sign and repulsive if the charges have the same sign; is a conservative force.

23.3 Coulomb’s Law From experimental observations on the electric force, we can express Coulomb’s law as an equation giving the magnitude of the electric force (sometimes called the Coulomb force) between two point charges: Where: r: is separation between the particles and directed along the line joining them

23.3 Coulomb’s Law q: is a charge ke is a constant called the Coulomb constant

23.3 Coulomb’s Law This constant is also written in the form The smallest unit of charge e is the charge on an electron (-e) or a proton (+e) and has a magnitude

23.3 Coulomb’s Law Example 23.1 The Hydrogen Atom The electron and proton of a hydrogen atom are separated(on the average) by a distance of approximately 5.3 x 1011 m. Find the magnitudes of the electric force

Coulomb Law Illustrated Like charges repel Unlike charges attract + r – + – If charges are of same magnitude (and same separation), all the forces will be the same magnitude, with different directions.

23.3 Coulomb’s Law

Coulomb Force Law, Qualitatively Double one of the charges force doubles Change sign of one of the charges force changes direction Change sign of both charges force stays the same Double the distance between charges force four times weaker Double both charges force four times stronger

23.3 Coulomb’s Law When more than two charges are present, the force between any pair of them is given by Equation Therefore, the resultant force on any one of them equals the vector sum of the forces exerted by the various individual charges. For example, if four charges are present, then the resultant force exerted by particles 2, 3, and 4 on particle 1 is

23.3 Coulomb’s Law Example 23.2 Find the Resultant Force Consider three point charges located at the corners of a right triangle as shown in Figure

23.3 Coulomb’s Law Solution : see the book

Coulomb’s Law: Ex 1 Determine the magnitude of the force between a proton and an electron in a hydrogen atom. Assume the distance from the electron to the nucleus is 0.53 X 10-10 m. + -

Coulomb’s Law: Ex 1 F = k Q1Q2 r2 F = 8.2 X 10-8 N F = (9.0 X 109 N-m2/C2 )(1.602 X 10-19 C)(1.602X10-19 C) (0.53 X 10-10 m)2 F = 8.2 X 10-8 N

Coulomb’s Law: Ex 2 What is the force between an electron and the three protons in a Li atom if the distance is about 1.3 X 10-10? (ANS: 3.9 X 10-8 N)

Coulomb’s Law: Ex 3 Three charged particles are arranged in a straight line as shown in the diagram. Calculate the net force on particle 3. 0.30 m 0.20 m - - + Q1= -8.0 mC Q2= +3.0 mC Q3 = -4.0 mC

Coulomb’s Law: Ex 3 F = k Q1Q2 r2 F31 = (9.0 X 109 N-m2/C2 )(4.0X 10-6 C)(8.0X10-6 C) (0.50 m)2 F31 = 1.2 N (Repulsive to the right) F32 = (9.0 X 109 N-m2/C2 )(4.0X 10-6 C)(3.0X10-6 C) (0.20 m)2 F32 = 2.7 N (Attractive to the left)

Fnet = F31 - F32 Fnet = 1.2 N – 2.7 N = -1.5 to the left

Coulomb’s Law: Ex 4 What is the resultant force on charge q3 if the charges are arranged as shown below. The magnitudes of the charges are: q1 = +6.00 X 10-9 C q2 = -2.00 X 10-9 C q3 = +5.00 X 10-9 C

Coulomb’s Law: Ex 4 4.00 m q2 - + q3 37o 3.00 m 5.00 m q1 +

Coulomb’s Law: Ex 4 First calculate the forces on q3 separately: F13= k Q1Q3 r2 F13 = (9.0 X 109 N-m2/C2)(6.00 X 10-9 C)(5.00 X 10-9 C) (5.00m )2 F13 = 1.08 X 10-8 N

Coulomb’s Law: Ex 4 F23 = k Q2Q3 r2 (4.00m )2 F23 = (9.0 X 109 N-m2/C2)(2.00 X 10-9 C)(5.00 X 10-9 C) (4.00m )2 F23 = 5.62 X 10-9 N

F13 = 1.08 X 10-8 N F23 = 5.62 X 10-9 N 37o q2 - + q3 37o q1 +

Coulomb’s Law: Ex 4 F13x = F13cos37o = (1.08 X 10-8 N)cos37o F13x = 8.63 X 10-9 N F13y = F13sin37o = (1.08 X 10-8 N)sin37o F13y = 6.50 X 10-9 N Fx = F23 + F13x Fx = -5.62 X 10-9 N + 8.63 X 10-9 N = 3.01 X 10-9 N Fy = F13y = 6.50 X 10-9 N

Coulomb’s Law: Ex 4 - q FR = \/ (3.01 X 10-9 N)2 + (6.50 X 10-9 N)2 Fx = 3.01 X 10-9 N Fy = 6.50 X 10-9 N FR = \/ (3.01 X 10-9 N)2 + (6.50 X 10-9 N)2 FR = 7.16 X 10-9 N sin q = Fy/FR sin q = (6.50 X 10-9 N)/ (7.16 X 10-9 N) q = 64.7o FR Fy q Fx -

Coulomb’s Law: Ex 5 Calculate the net electrostatic force on charge Q3 as shown in the diagram: Q3 = +65 mC + 60 cm 30 cm 30o + - Q2 = +50 mC Q1 = -86 mC

Coulomb’s Law: Ex 5 First calculate the forces on Q3 separately: F13= k Q1Q3 r2 F13 = (9.0 X 109 N-m2/C2)(65 X 10-6 C)(86 X 10-6 C) (0.60 m )2 F13 = 140 N F23 = (9.0 X 109 N-m2/C2)(65 X 10-6 C)(50 X 10-6 C) (0.30 m )2 F23 = 330 N

Coulomb’s Law: Ex 5 F23 Q3 + F13 30o + - Q2 Q1

Coulomb’s Law: Ex 5 F13x = F13cos30o = (140 N)cos30o F13x = 120 N F13y = -F13sin30o = (140 N)sin30o F13y = -70 N Fx = F13x = 120 N F7 = 330 N - 70 N = 260 N

Coulomb’s Law: Ex 5 - q FR = \/ (120 N)2 + (330 N)2 FR = 290 N Fx = 120 N Fy = 260 N FR = \/ (120 N)2 + (330 N)2 FR = 290 N sin q = Fy/FR sin q = (260 N)/ (290 N) q = 64o FR Fy q Fx -